Question

There are 15 tennis balls in a box, of which 9 have not previously been used. Three of the balls are randomly chosen, played with, and then returned to the box. Later, another 3 balls are randomly chosen from the box. Find the probability that none of these balls has ever been used.

Short answer

The likelihood is that none of these balls have ever been used is$.0893$

Step-by-step solution

Step1:Given data

Case $0$: No used balls are drawn. $p_0=\frac{\left(\begin{array}{c}9\\ 3\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}$

Case $1:1$used ball is drawn. $p_1=\frac{\left(\begin{array}{c}9\\ 2\end{array}\right)·6}{\left(\begin{array}{c}15\\ 3\end{array}\right)}$

Case$2:2$ used balls are drawn.$p_2=\frac{9·\left(\begin{array}{c}6\\ 2\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}$

Case$3:3$ used balls are drawn. $p_3=\frac{\left(\begin{array}{c}6\\ 3\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}$

Step2:Three of the balls are chosen at random.

Case $0:p_0=.1846,6$new balls, $9$used left over.

Case $1:p_1=.4747,7$new balls,$8$used.

Case$2:p_2=.2967,8$new balls,$7$used.

Case$3:p3=.044,9$ new balls, $6$ used.

Another 3 ball is drawn at random from the box.

$p_0\frac{\left(\begin{array}{c}6\\ 3\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}+p_1\frac{\left(\begin{array}{l}7\\ 3\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}+p_2\frac{\left(\begin{array}{l}8\\ 3\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}+p_3\frac{\left(\begin{array}{l}9\\ 3\end{array}\right)}{\left(\begin{array}{c}15\\ 3\end{array}\right)}$

We multiply each case's probability by the probability that no used balls are found in the second draw, and then add it all up.

Step4:It's possible that none of these balls have ever been used.

$$\begin{gathered} .1846\times .044+.4747\times .0769+.2967\times .1231+.044\times .1846 \\ =.0893 \end{gathered}$$